INVERSE TRIGONOMETRY MCQs for JEE MATHEMATICS — Mixed difficulty

1 Medium

Application +4 | -1

The value of

\cos ^{-1}(\cos 12)-\sin ^{-1}(\sin 12)

is-

View answer and explanation

Correct option: C

8 \pi-4

Option C (

8 \pi - 24

) is correct because it involves finding the correct range for the inverse trigonometric functions and then subtracting them.

Solution approach

Correct Answer + Core Reason: $8 \pi-24$ . The value of $\cos^{-1}(\cos 12)-\sin^{-1}(\sin 12)$ is found by adjusting the angles to be within the principal ranges of the inverse trigonometric functions.

Step-by-Step Solution:

Step 1: We are asked to find $\cos^{-1}(\cos 12) - \sin^{-1}(\sin 12)$ .

Step 2: The principal range of $\cos^{-1}(x)$ is $[0, \pi]$ and the principal range of $\sin^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$

2 Medium

2 of 581Question permalink

Application +4 | -1

The value of

\sin \left(2 \tan ^{-1}(1 / 3)\right)+

\cos \left(\tan ^{-1} 2 \sqrt{2}\right)

is

View answer and explanation

Correct option: C

14 / 15

Option C (14/15) is correct because by evaluating the sine and cosine terms using trigonometric identities and substituting, we arrive at the correct answer.

Solution approach

Correct Answer + Core Reason: $14 / 15$ . Evaluating the sine and cosine terms.

Step-by-Step Solution:

Step 1: Given $\sin \left(2 \tan^{-1} (1/3)\right) + \cos \left(\tan^{-1} 2\sqrt{2}\right)$ .

Step 2: Let $\alpha = \tan^{-1} (1/3)$ . Then $\tan \alpha = 1/3$ . We want to find $\sin 2\alpha$ .

Step 3: $\sin 2\alpha = \frac{2 \tan \alpha}{1 + \tan^2 \alpha} = \frac{2(1/3)}{1 + (1/9)} = \frac{2/3}{10/9} = \frac{2}{3} \cdot \frac{9}{10} = \frac{3}{5}$ .

Step 4: Let $\beta = \tan^{-1} 2\sqrt{2}$ . Then $\tan \beta = 2\sqrt{2}$ . We want to find $\cos \beta$ .

Step 5: $\cos \beta = \frac{1}{\sqrt{1 + \tan^2 \beta}} = \frac{1}{\sqrt{1 + (2\sqrt{2})^2}} = \frac{1}{\sqrt{1 + 8}} = \frac{1}{\sqrt{9}} = \frac{1}{3}$ .

Step 6: Therefore, $\sin \left(2 \tan^{-1} (1/3)\right) + \cos \left(\tan^{-1} 2\sqrt{2}\right) = \frac{3}{5} + \frac{1}{3} = \frac{9 + 5}{15} = \frac{14}{15}$ .

Why other options are wrong:
- Option A: 12/13 is incorrect. It does not follow from the correct calculations.
- Option B: 13/14 is incorrect. It does not follow from the correct calculations.
- Option D: none of these is incorrect because 14/15 is the correct answer.

Key Concepts:
- Inverse trigonometric functions
- Trigonometric identities

Memory Aids:
- Remember the formula for $\sin 2\alpha$ .

Quick Tips:
- When dealing with inverse trigonometric functions, try to simplify the expression using trigonometric identities.

3 Medium

3 of 581Question permalink

Application +4 | -1

Find

\cos ^ {-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right\}=

View answer and explanation

Correct option: D

\frac{17 \pi}{20}

Option D (

\frac{{17 \pi}}{{20}}

) is correct because it simplifies the given expression using trigonometric identities and inverse cosine properties.

Solution approach

Correct Answer + Core Reason: $\frac{{17 \pi}}{{20}}$ . The expression simplifies to $\cos^{-1}(\cos(\frac{{17\pi}}{{20}}))$ which equals $\frac{{17\pi}}{{20}}$ since it lies within the range of inverse cosine.

Step-by-Step Solution:

Step 1: Rewrite the expression inside the inverse cosine:
$\frac{{1}}{{\sqrt{{2}}}}(\cos \frac{{9 \pi}}{{10}} - \sin \frac{{9 \pi}}{{10}}) = \cos \frac{{\pi}}{{4}} \cos \frac{{9 \pi}}{{10}} - \sin \frac{{\pi}}{{4}} \sin \frac{{9 \pi}}{{10}}$

Step 2: Apply the cosine addition formula: $\cos(A + B) = \cos A \cos B - \sin A \sin B$ \nSo, the expression becomes: $\cos(\frac{{\pi}}{{4}} + \frac{{9 \pi}}{{10}}) = \cos(\frac{{5\pi + 18\pi}}{{20}}) = \cos(\frac{{23 \pi}}{{20}})$

Step 3: Substitute back into the original expression:
$\cos^{-1}(\cos(\frac{{23 \pi}}{{20}}))$

Step 4: Since $\frac{{23 \pi}}{{20}}$ is not within the range of $\cos^{-1}(x)$ which is $[0, \pi]$ , we need to find an equivalent angle within this range.
$\frac{{23π}}{{20}}$ = 2π - $\frac{{17π}}{{20}}$ $\n$ \cos(2π - θ) = \cos(θ) $\nSo,$ \cos( $\frac{{23 π}}{{20}}$ ) = \cos( $\frac{{17 π}}{{20}}$ ) $\n\n**Step 5:** Therefore,$ \cos^-1(\cos( $\frac{{23 π}}{{20}}$ )) = \cos^-1(\cos( $\frac{{17 π}}{{20}}$ )) = $\frac{{17 π}}{{20}}$ $\n\n**Why other options are wrong:**\n- **Option A:**$ $\frac{{3π}}{{20}}$ $: This is incorrect as it doesn't account for the correct simplification and range of inverse cosine.\n- **Option B:**$ $\frac{{7π}}{{20}}$ $: This is incorrect as it doesn't account for the correct simplification and range of inverse cosine.\n- **Option C:**$ $\frac{{7π}}{{10}}$ $: This is incorrect as it doesn't account for the correct simplification and range of inverse cosine.\n\n**Key Concepts:**\n- Trigonometric Identities (Cosine Addition Formula)\n- Inverse Trigonometric Functions (Range of Inverse Cosine)\n\n**Memory Aids:**\n- Cosine Addition Formula:$ \cos(A + B) = \cos A \cos B - \sin A \sin B $\n- Range of$ \cos^-1(x) $is$ [0, π] $\n\n**Quick Tips:**\n- Always check if the angle is within the range of the inverse trigonometric function.\n- Use trigonometric identities to simplify the expression before evaluating the inverse function.$

4 Hard

4 of 581Question permalink

Application +4 | -1

If

\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1

, then

x

is equal to

View answer and explanation

Correct option: D

\frac{1}{5}

Option D (

\frac{1}{5}

) is correct because solving the equation

\sin(\sin^{-1} \frac{1}{5} + \cos^{-1} x) = 1

leads to

x = \frac{1}{5}

.

Solution approach

Correct Answer + Core Reason: $\frac{1}{5}$ . Solving the equation $\sin(\sin^{-1} \frac{1}{5} + \cos^{-1} x) = 1$ leads to $x = \frac{1}{5}$ .

Step-by-Step Solution:

Step 1: Given $\sin(\sin^{-1} \frac{1}{5} + \cos^{-1} x) = 1$ .

Step 2: Take the inverse sine of both sides.
$\sin^{-1} \frac{1}{5} + \cos^{-1} x = \sin^{-1} 1 = \frac{\pi}{2}$ .

Step 3: Rearrange the equation.
$\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} \frac{1}{5}$ .

Step 4: Use the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ .
$\cos^{-1} x = \cos^{-1} \frac{1}{5}$ .

Step 5: Take the cosine of both sides.\nx = $\frac{1}{5}$ .

Why other options are wrong:
- Option A (1): This is incorrect because it does not satisfy the original equation.
- Option B (-1): This is incorrect because it does not satisfy the original equation.
- **Option C ( $\frac{4}{5}$ ): This is incorrect because it does not satisfy the original equation.

Key Concepts:**
- Inverse trigonometric functions
- Trigonometric identities
- $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$

Memory Aids:
- Remember the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ .

Quick Tips:
- Use the identities to simplify trigonometric expressions.

5 Easy

5 of 581Question permalink

Application +4 | -1

If

{x^2} + {y^2} + {z^2} = {r^2},

then

{\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{xz}}{{yr}}} \right)\;

is equal to

View answer and explanation

Correct option: B

\frac{\pi }{2}

Option B (

\frac{\pi }{2}

) is correct because the expression simplifies to

\tan^{-1}(\infty)

, which equals

\frac{\pi }{2}

given the condition

{x^2} + {y^2} + {z^2} = {r^2}

.

Solution approach

Correct Answer + Core Reason: $\frac{\pi }{2}$ . The expression simplifies to $\tan^{-1}(\infty)$ , which equals $\frac{\pi }{2}$ given the condition ${x^2} + {y^2} + {z^2} = {r^2}$ .

Step-by-Step Solution:

Step 1: Given ${x^2} + {y^2} + {z^2} = {r^2}$ .

Step 2: Consider the expression ${\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{xz}}{{yr}}} \right)$ .

Step 3: Use the formula ${\tan ^{ - 1}}a + {\tan ^{ - 1}}b + {\tan ^{ - 1}}c = {\tan ^{ - 1}}\left[ {\frac{{a + b + c - abc}}{{1 - (ab + bc + ca)}}} \right]$ .\nLet $a = \frac{{xy}}{{zr}}$ , $b = \frac{{yz}}{{xr}}$ , and $c = \frac{{xz}}{{yr}}$ .\nThen, $a + b + c = \frac{{xy}}{{zr}} + \frac{{yz}}{{xr}} + \frac{{xz}}{{yr}}$ and $abc = \frac{{xy}}{{zr}} \cdot \frac{{yz}}{{xr}} \cdot \frac{{xz}}{{yr}} = \frac{{{x^2}{y^2}{z^2}}}{{{x^2}{y^2}{z^2}{r^3}}} = \frac{{xyz}}{{{r^3}}}$ .\nAlso, $ab = \frac{{xy}}{{zr}} \cdot \frac{{yz}}{{xr}} = \frac{{{y^2}}}{{{r^2}}}$ , $bc = \frac{{yz}}{{xr}} \cdot \frac{{xz}}{{yr}} = \frac{{{z^2}}}{{{r^2}}}$ , and $ca = \frac{{xz}}{{yr}} \cdot \frac{{xy}}{{zr}} = \frac{{{x^2}}}{{{r^2}}}$ .\nSo, $ab + bc + ca = \frac{{{y^2}}}{{{r^2}}} + \frac{{{z^2}}}{{{r^2}}} + \frac{{{x^2}}}{{{r^2}}} = \frac{{{x^2} + {y^2} + {z^2}}}{{{r^2}}}$ .

Step 4: Substitute into the formula:
${\tan ^{ - 1}}\left[ {\frac{{\frac{{xy}}{{zr}} + \frac{{yz}}{{xr}} + \frac{{xz}}{{yr}} - \frac{{xyz}}{{{r^3}}}}}}{{1 - \frac{{{x^2} + {y^2} + {z^2}}}{{{r^2}}}}}} \right]$ .\nSince ${x^2} + {y^2} + {z^2} = {r^2}$ , the denominator becomes $1 - \frac{{{r^2}}}{{{r^2}}} = 1 - 1 = 0$ .

Step 5: The expression becomes ${\tan ^{ - 1}}\left[ {\frac{{\frac{{xy}}{{zr}} + \frac{{yz}}{{xr}} + \frac{{xz}}{{yr}} - \frac{{xyz}}{{{r^3}}}}}{0}} \right] = {\tan ^{ - 1}}(\infty)$ .

Step 6: Evaluate ${\tan ^{ - 1}}(\infty)$ .
${\tan ^{ - 1}}(\infty) = \frac{\pi }{2}$ .

Why other options are wrong:
- Option A: $\pi$ is incorrect because the expression does not simplify to ${\tan ^{ - 1}}(0)$ .
- Option C: 0 is incorrect because the expression does not simplify to ${\tan ^{ - 1}}(0)$ .
- Option D: None of these is incorrect because $\frac{\pi }{2}$ is the correct answer.

Key Concepts:
- Inverse trigonometric functions
- Sum of inverse tangents formula

Memory Aids:
- Remember the formula: ${\tan ^{ - 1}}a + {\tan ^{ - 1}}b + {\tan ^{ - 1}}c = {\tan ^{ - 1}}\left[ {\frac{{a + b + c - abc}}{{1 - (ab + bc + ca)}}} \right]$

6 Easy

6 of 581Question permalink

Application +4 | -1

2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)

is equal to

View answer and explanation

Correct option: D

\frac{\pi }{4}

Option D (

\frac{\pi }{4}

) is correct because it is the result of applying the tangent addition formula twice to simplify the expression.

Solution approach

Correct Answer + Core Reason: $\frac{\pi }{4}$ . Applying the tangent addition formula twice simplifies the expression to $\tan^{-1}(1)$ .

Step-by-Step Solution:

Step 1: Given $2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)$ .

Step 2: Using the formula $2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)$ , we have $2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) = {\tan ^{ - 1}}\left( {\frac{{2\left( {\frac{1}{3}} \right)}}{{1 - \frac{1}{9}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{\frac{2}{3}}}{{\frac{8}{9}}}} \right) = {\tan ^{ - 1}}\left( {\frac{3}{4}} \right)$ .

Step 3: Now, we have ${\tan ^{ - 1}}\left( {\frac{3}{4}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)$ .

Step 4: Using the formula ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right)$ , we have ${\tan ^{ - 1}}\left( {\frac{3}{4}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right) = {\tan ^{ - 1}}\left( {\frac{{\frac{3}{4} + \frac{1}{7}}}{{1 - \frac{3}{4} \times \frac{1}{7}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{\frac{{21 + 4}}{{28}}}}{{\frac{{28 - 3}}{{28}}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{25}}{{25}}} \right) = {\tan ^{ - 1}}(1)$ .

Step 5: Since ${\tan ^{ - 1}}(1) = \frac{\pi }{4}$ , the final answer is $\frac{\pi }{4}$ .

Why other options are wrong:
- Option A: $\left( {\frac{{49}}{{29}}} \right)$ is incorrect because it is not the correct simplification of the given expression.
- Option B: $\frac{\pi }{2}$ is incorrect because it would imply that the tangent of the expression is infinite, which is not the case.
- Option C: $- \left( {\frac{{49}}{{29}}} \right)$ is incorrect because it is a negative value, while the expression evaluates to a positive angle.

Key Concepts:
- Inverse trigonometric functions
- Tangent addition formula

Memory Aids:
- Remember the tangent addition formula: ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right)$ .

Quick Tips:
- Simplify the expression step by step using the appropriate formulas.

7 Medium

7 of 581Question permalink

Application +4 | -1

The number of positive integral solutions of the equation

\tan ^{-1} x+\cot ^{-1} y=\tan ^{-1} 4

is

View answer and explanation

Correct option: B

1

Option B (1) is correct because by simplifying the equation, we find only one positive integral solution that satisfies the given condition.

Solution approach

Correct Answer + Core Reason: 1. Simplifying the equation leads to only one positive integral solution.

Step-by-Step Solution:

Step 1: $\tan ^{-1} x+\cot ^{-1} y=\tan ^{-1} 4 \Rightarrow \tan^{-1} x - \tan^{-1} (1/y) = \tan^{-1} 4$

Step 2: $\tan ^{-1} x-\tan ^{-1} y=-\left(\frac{\pi}{2}-\tan ^{-1} 4\right)$ is incorrect. The correct approach is $\tan^{-1} x + \cot^{-1} y = \tan^{-1} x + \tan^{-1}(1/y) = \tan^{-1} 4$ .

Step 3: $\tan^{-1} \frac{x + 1/y}{1 - x/y} = \tan^{-1} 4$ .

Step 4: $\frac{x + 1/y}{1 - x/y} = 4 \Rightarrow \frac{xy + 1}{y - x} = 4 \Rightarrow xy + 1 = 4y - 4x \Rightarrow xy - 4y = -4x - 1 \Rightarrow y(x - 4) = -4x - 1 \Rightarrow y = \frac{-4x - 1}{x - 4} = \frac{4x + 1}{4 - x}$ .

Step 5: Since $x$ and $y$ are positive integers, $4 - x > 0$ , so $x < 4$ . Thus, $x$ can be 1, 2, or 3.

Step 6: If $x = 1$ , $y = \frac{4(1) + 1}{4 - 1} = \frac{5}{3}$ , which is not an integer.

Step 7: If $x = 2$ , $y = \frac{4(2) + 1}{4 - 2} = \frac{9}{2}$ , which is not an integer.

Step 8: If $x = 3$ , $y = \frac{4(3) + 1}{4 - 3} = \frac{13}{1} = 13$ , which is an integer.

Step 9: Therefore, the only positive integral solution is $(3, 13)$ .

Why other options are wrong:
- Option A: This option is incorrect as there is one solution.
- Option C: This option is incorrect as there is only one solution.
- Option D: This option is incorrect as there is only one solution.

Key Concepts:
- Inverse trigonometric functions
- Tangent addition formula

Memory Aids:
- Remember the tangent addition formula.

Quick Tips:
- Convert all inverse trigonometric functions to the same function before solving.

8 Medium

8 of 581Question permalink

Application +4 | -1

If

{\sin ^{ - 1}}\frac{{2a}}{{1 + {a^2}}} - {\cos ^{ - 1}}\frac{{1 - {b^2}}}{{1 + {b^2}}} = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}}

, then the value of

x

is

View answer and explanation

Correct option: D

\frac{{a - b}}{{1 + ab}}

Option D (

\frac{{a - b}}{{1 + ab}}

) is correct because it is derived by using trigonometric identities to simplify the given equation and solve for

x

.

Solution approach

Correct Answer + Core Reason: $\frac{{a - b}}{{1 + ab}}$ . The value of x is found by using trigonometric identities.

Step-by-Step Solution:

Step 1: Recognize the trigonometric identities: ${\sin ^{ - 1}}\frac{{2a}}{{1 + {a^2}}} = 2{\tan ^{ - 1}}a$ , ${\cos ^{ - 1}}\frac{{1 - {b^2}}}{{1 + {b^2}}} = 2{\tan ^{ - 1}}b$ , and ${\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} = 2{\tan ^{ - 1}}x$ .

Step 2: Substitute these identities into the given equation: $2{\tan ^{ - 1}}a - 2{\tan ^{ - 1}}b = 2{\tan ^{ - 1}}x$ .

Step 3: Divide both sides by 2: ${\tan ^{ - 1}}a - {\tan ^{ - 1}}b = {\tan ^{ - 1}}x$ .

Step 4: Apply the arctangent subtraction formula: ${\tan ^{ - 1}}a - {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left( {\frac{{a - b}}{{1 + ab}}} \right)$ .

Step 5: Equate the arguments of the arctangent functions: $x = \frac{{a - b}}{{1 + ab}}$ .

Why other options are wrong:
- Option A: $a$ is incorrect because it doesn't account for the subtraction of the second term.
- Option B: $b$ is incorrect because it doesn't account for the first term and the relationship between $a$ and $b$ .
- Option C: $\frac{{a + b}}{{1 - ab}}$ is incorrect because it represents the arctangent addition formula, not the subtraction formula.

Key Concepts:
- Trigonometric identities: ${\sin ^{ - 1}}\frac{{2a}}{{1 + {a^2}}} = 2{\tan ^{ - 1}}a$ , ${\cos ^{ - 1}}\frac{{1 - {b^2}}}{{1 + {b^2}}} = 2{\tan ^{ - 1}}b$ , ${\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} = 2{\tan ^{ - 1}}x$ .
- Arctangent subtraction formula: ${\tan ^{ - 1}}a - {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left( {\frac{{a - b}}{{1 + ab}}} \right)$ .

Memory Aids:
- Remember the trigonometric identities and the arctangent subtraction formula.

Quick Tips:
- Recognize the trigonometric identities to simplify the equation.

9 Medium

9 of 581Question permalink

Application +4 | -1

The value of

\frac{\alpha^2}{2} \operatorname{cosec}^2\left(\frac{1}{2} \tan ^{-1} \frac{\alpha}{\beta}\right)-\frac{\beta^2}{2} \sec ^2\left(\frac{1}{2} \tan ^{-1} \frac{\beta}{\alpha}\right)

is equal to

View answer and explanation

Correct option: D

\sqrt{\alpha^2+\beta^2}(\beta+\alpha)

Option D (

\sqrt{{\alpha^2+\beta^2}}(\beta+\alpha)

) is correct because after simplifying the trigonometric expressions and substituting, the value equals to the given expression.

Solution approach

Correct Answer + Core Reason: $\sqrt{{\alpha^2+\beta^2}}(\beta+\alpha)$ . After simplifying the trigonometric expressions and substituting, the value equals to the given expression.

Step-by-Step Solution:

Step 1: Given expression is $\frac{{\alpha^2}}{{2}} \operatorname{{cosec}}^2(\frac{{1}}{{2}} \tan ^{-1} \frac{{\alpha}}{{\beta}}) - \frac{{\beta^2}}{{2}} \sec ^2(\frac{{1}}{{2}} \tan ^{-1} \frac{{\beta}}{{\alpha}})$ .

Step 2: Using the identities $\operatorname{{cosec}}^2(\frac{{\theta}}{{2}}) = \frac{{2}}{{1 - \cos \theta}}$ and $\sec ^2(\frac{{\theta}}{{2}}) = \frac{{2}}{{1 + \cos \theta}}$ , the expression becomes $\alpha^2 \frac{{1}}{{1 - \cos(\tan ^{-1} \frac{{\alpha}}{{\beta}})}} - \beta^2 \frac{{1}}{{1 + \cos(\tan ^{-1} \frac{{\beta}}{{\alpha}})}}$ .

Step 3: Let $\tan ^{-1} \frac{{\alpha}}{{\beta}} = x$ , so $\tan x = \frac{{\alpha}}{{\beta}}$ . Then $\cos x = \frac{{\beta}}{{\sqrt{{\alpha^2 + \beta^2}}}}$ and $\tan ^{-1} \frac{{\beta}}{{\alpha}} = y$ , so $\tan y = \frac{{\beta}}{{\alpha}}$ . Then $\cos y = \frac{{\alpha}}{{\sqrt{{\alpha^2 + \beta^2}}}}$

Step 4: Substituting these values, we get $\frac{{\alpha^2}}{{1 - \frac{{\beta}}{{\sqrt{{\alpha^2 + \beta^2}}}}}} - \frac{{\beta^2}}{{1 + \frac{{\alpha}}{{\sqrt{{\alpha^2 + \beta^2}}}}}} = \frac{{\alpha^2 \sqrt{{\alpha^2 + \beta^2}}}}{{\sqrt{{\alpha^2 + \beta^2}} - \beta}} - \frac{{\beta^2 \sqrt{{\alpha^2 + \beta^2}}}}{{\sqrt{{\alpha^2 + \beta^2}} + \alpha}}$ .

Step 5: Rationalizing the denominators, we get $\sqrt{{\alpha^2 + \beta^2}} [\frac{{\alpha^2(\sqrt{{\alpha^2 + \beta^2}} + \beta)}}{{\alpha^2}} - \frac{{\beta^2(\sqrt{{\alpha^2 + \beta^2}} - \alpha)}}{{\beta^2}}]$

10 Medium

10 of 581Question permalink

Application +4 | -1

2\left(\tan ^{-1} 1+\tan ^{-1} 2+\tan ^{-1} 3\right)

is equal to

View answer and explanation

Correct option: D

2 \pi

Option D (

2 \pi

) is correct because the expression

2(\tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3)

simplifies to

2\pi

using the properties of inverse tangent functions.

Solution approach

Correct Answer + Core Reason: $2 \pi$ . The expression simplifies to $2\pi$ using inverse tangent properties.

Step-by-Step Solution:

Step 1: Consider $\tan^{-1} 2 + \tan^{-1} 3$ .

Step 2: Use the formula $\tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1} \frac{x+y}{1-xy}$ when $x > 0$ , $y > 0$ , and $xy > 1$ .

Step 3: Apply the formula: $\tan^{-1} 2 + \tan^{-1} 3 = \pi + \tan^{-1} \frac{2+3}{1-(2)(3)} = \pi + \tan^{-1} \frac{5}{-5} = \pi + \tan^{-1} (-1)$ .

Step 4: Since $\tan^{-1} (-1) = -\frac{\pi}{4}$ , we have $\tan^{-1} 2 + \tan^{-1} 3 = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$ . However, since $x, y > 0$ and $xy > 1$ , we use $\pi + \tan^{-1}(\frac{x+y}{1-xy})$ . So, $\tan^{-1} 2 + \tan^{-1} 3 = \pi + \tan^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$ . This is incorrect. The correct application is $\tan^{-1} 2 + \tan^{-1} 3 = \pi + \tan^{-1}(\frac{2+3}{1-2*3}) = \pi + \tan^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$ .

Step 5: Now, consider $\tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3 = \frac{\pi}{4} + \pi + \tan^{-1}(\frac{5}{-5}) = \frac{\pi}{4} + \pi - \frac{\pi}{4} = \pi$ .

Step 6: Therefore, $2(\tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3) = 2\pi$ .

Step 7: Final Answer: $2\pi$

Why other options are wrong:
- Option A: This is incorrect because it does not account for the properties of inverse tangent functions when $xy > 1$ .
- Option B: This is incorrect because it does not account for the properties of inverse tangent functions when $xy > 1$ .
- Option C: This is incorrect because it does not account for the factor of 2.

Key Concepts:
- Inverse trigonometric functions
- Addition formula for inverse tangent

Memory Aids:
- Remember the addition formula: $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \frac{x+y}{1-xy}$ (with adjustments for different quadrants).

Quick Tips:
- Be careful when using the addition formula for inverse tangent, especially when $xy > 1$ . Remember to add or subtract $\pi$ accordingly.

11 Easy

11 of 581Question permalink

Analyzing +4 | -1

For the equation

{\cos ^{ - 1}}x + {\cos ^{ - 1}}2x + \pi = 0

, then the number of real solutions is

View answer and explanation

Correct option: C

0

Option C (0) is correct because the equation has no real solutions, as the value of x obtained does not satisfy the original equation.

Solution approach

Correct Answer + Core Reason: 0. The equation has no real solutions, as the value of x obtained does not satisfy the original equation.

Step-by-Step Solution:

Step 1: Rewrite the given equation.
${\cos ^{ - 1}}x + {\cos ^{ - 1}}2x + \pi = 0$
${\cos ^{ - 1}}x + {\cos ^{ - 1}}2x = - \pi$

Step 2: Apply the formula ${\cos ^{ - 1}}A + {\cos ^{ - 1}}B = {\cos ^{ - 1}}(AB - \sqrt {1 - {A^2}} \sqrt {1 - {B^2}} )$ .
${\cos ^{ - 1}}\left( {x \cdot 2x - \sqrt {1 - {x^2}} \sqrt {1 - 4{x^2}} } \right) = - \pi$

Step 3: Take the cosine of both sides.
$2{x^2} - \sqrt {1 - {x^2}} \sqrt {1 - 4{x^2}} = \cos ( - \pi ) = -1$

Step 4: Rearrange the equation.
$2{x^2} + 1 = \sqrt {1 - {x^2}} \sqrt {1 - 4{x^2}}$

Step 5: Square both sides.
${\left( {2{x^2} + 1} \right)^2} = \left( {1 - {x^2}} \right)\left( {1 - 4{x^2}} \right)$
$4{x^4} + 4{x^2} + 1 = 1 - 5{x^2} + 4{x^4}$

Step 6: Simplify the equation.
$9{x^2} = 0$
$x = 0$

Step 7: Check if the solution satisfies the original equation.
${\cos ^{ - 1}}(0) + {\cos ^{ - 1}}(0) + \pi = \frac{\pi }{2} + \frac{\pi }{2} + \pi = 2\pi \ne 0$ \nTherefore, $x = 0$ is not a solution.

Why other options are wrong:
- Option A: 1 is incorrect because there are no real solutions.
- Option B: 2 is incorrect because there are no real solutions.
- Option D: $\infty$ is incorrect because there are no real solutions.

Key Concepts:
- Inverse trigonometric functions
- Solving trigonometric equations

Memory Aids:
- Remember the range of ${\cos ^{ - 1}}x$ is $[0, \pi]$ .

Quick Tips:
- Always check the solutions in the original equation to avoid extraneous solutions.

12 Easy

12 of 581Question permalink

Application +4 | -1

If

{\sin ^{ - 1}}a + {\sin ^{ - 1}}b + {\sin ^{ - 1}}c = \pi ,

then the value of

a\sqrt {\left( {1 - {a^2}} \right)} + b\sqrt {\left( {1 - {b^2}} \right)} + c\sqrt {1 - {c^2})} \;

will be

View answer and explanation

Correct option: A

2

abc

Option A (2

abc

) is correct because it is derived from the trigonometric identity

\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C

when

A + B + C = \pi

, where

A = {\sin ^{ - 1}}a

,

B = {\sin ^{ - 1}}b

, and

C = {\sin ^{ - 1}}c

.

Solution approach

Correct Answer + Core Reason: 2 $abc$ . The value is obtained by using the trigonometric identity $\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C$ when $A + B + C = \pi$ .

Step-by-Step Solution:

Step 1: Let ${\sin ^{ - 1}}a = A,{\sin ^{ - 1}}b = B,{\sin ^{ - 1}}c = C$ . Then $\sin A = a, \sin B = b, \sin C = c$ .

Step 2: Given $A + B + C = \pi$ .

Step 3: Use the identity $\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C$ .

Step 4: Rewrite the identity as $\sin A\cos A + \sin B\cos B + \sin C\cos C = 2\sin A\sin B\sin C$ .

Step 5: Substitute $\sin A = a, \sin B = b, \sin C = c$ and $\cos A = \sqrt{1 - a^2}, \cos B = \sqrt{1 - b^2}, \cos C = \sqrt{1 - c^2}$ .

Step 6: $a\sqrt {(1 - {a^2})} + b\sqrt {(1 - {b^2})} + c\sqrt {1 - {c^2}} = 2abc$ .

Why other options are wrong:
- Option B: $abc$ : This is incorrect because it is missing the factor of 2.
- Option C: $\frac{1}{2}abc$ : This is incorrect because the factor should be 2, not $\frac{1}{2}$ .
- Option D: $\frac{1}{3}abc$ : This is incorrect because the factor should be 2, not $\frac{1}{3}$ .

Key Concepts:
- Inverse trigonometric functions
- Trigonometric identities

Memory Aids:
- Remember the identity $\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C$ when $A + B + C = \pi$ .

Quick Tips:
- When dealing with inverse trigonometric functions, try to convert them to trigonometric functions and use identities.

13 Medium

13 of 581Question permalink

Application +4 | -1

\sin ^{-1} \frac{x}{5}+\operatorname{cosec}^{-1} \frac{5}{4}=\frac{\pi}{2}

then

x=

View answer and explanation

Correct option: D

3

Option D (3) is correct because using the identity

\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}

and

\operatorname{cosec}^{-1} x = \sin^{-1} \frac{1}{x}

, we can solve for x.

Solution approach

Correct Answer + Core Reason: 3. Using the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ and $\operatorname{cosec}^{-1} x = \sin^{-1} \frac{1}{x}$ , we can solve for x.

Step-by-Step Solution:

Step 1: Rewrite the cosecant inverse function as a sine inverse function.
$\operatorname{cosec}^{-1} \frac{5}{4} = \sin^{-1} \frac{4}{5}$ .

Step 2: Substitute this into the given equation.
$\sin^{-1} \frac{x}{5} + \sin^{-1} \frac{4}{5} = \frac{\pi}{2}$ .

Step 3: Use the identity $\sin^{-1} a + \cos^{-1} a = \frac{\pi}{2}$ .
$\sin^{-1} \frac{x}{5} = \frac{\pi}{2} - \sin^{-1} \frac{4}{5}$ .
$\sin^{-1} \frac{x}{5} = \cos^{-1} \frac{4}{5}$ .

Step 4: Take the sine of both sides.
$\sin(\sin^{-1} \frac{x}{5}) = \sin(\cos^{-1} \frac{4}{5})$ .
$\frac{x}{5} = \sin(\cos^{-1} \frac{4}{5})$ .

Step 5: Let $\theta = \cos^{-1} \frac{4}{5}$ . Then $\cos \theta = \frac{4}{5}$ .\nWe need to find $\sin \theta$ . Since $\sin^2 \theta + \cos^2 \theta = 1$ , we have $\sin^2 \theta = 1 - \cos^2 \theta = 1 - (\frac{4}{5})^2 = 1 - \frac{16}{25} = \frac{9}{25}$ .\nSo, $\sin \theta = \sqrt{\frac{9}{25}} = \frac{3}{5}$ .

Step 6: Substitute back into the equation.
$\frac{x}{5} = \frac{3}{5}$ .

Step 7: Solve for x.\nx = 3.

Final Answer: 3

Why other options are wrong:
- Option A (4): This is incorrect because substituting x=4 into the original equation does not satisfy the equation.
- Option B (5): This is incorrect because substituting x=5 into the original equation does not satisfy the equation.
- Option C (1): This is incorrect because substituting x=1 into the original equation does not satisfy the equation.

Key Concepts:
- Inverse trigonometric functions
- Identity: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$
- Identity: $\operatorname{cosec}^{-1} x = \sin^{-1} \frac{1}{x}$

Memory Aids:
- Remember the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ .
- Remember the identity $\operatorname{cosec}^{-1} x = \sin^{-1} \frac{1}{x}$ .

Quick Tips:
- Convert all inverse trigonometric functions to sine inverse or cosine inverse functions.
- Use the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ to simplify the equation.
- Solve the resulting equation for x.

14 Easy

14 of 581Question permalink

Application +4 | -1

If

\tan ^{-1} \frac{\sqrt{1+x^2}-1}{x}=4

then

x=

View answer and explanation

Correct option: D

\tan 8

Option D (

\tan 8

) is correct because substituting

x = \tan \theta

simplifies the expression to

(1/2) \tan^{-1}(x) = 4

, leading to

x = \tan 8

.

Solution approach

Correct Answer + Core Reason: $\tan 8$ . The equation is solved by substituting $x = \tan \theta$ and simplifying the inverse trigonometric expression.

Step-by-Step Solution:

Step 1: Substitute $x = \tan \theta$ into the given equation:
$\tan^{-1} \frac{\sqrt{1+x^2}-1}{x} = \tan^{-1} \frac{\sqrt{1+\tan^2 \theta}-1}{\tan \theta}$

Step 2: Simplify the expression using trigonometric identities:
$= \tan^{-1} \frac{\sec \theta - 1}{\tan \theta} = \tan^{-1} \frac{\frac{1}{\cos \theta} - 1}{\frac{\sin \theta}{\cos \theta}} = \tan^{-1} \frac{1 - \cos \theta}{\sin \theta}$

Step 3: Use the half-angle formulas:
$= \tan^{-1} \frac{2\sin^2 (\theta/2)}{2\sin(\theta/2)\cos(\theta/2)} = \tan^{-1} (\tan (\theta/2)) = \frac{1}{2} \theta$

Step 4: Substitute back $x = \tan \theta$ , so $\theta = \tan^{-1} x$ :
$\frac{1}{2} \tan^{-1} x = 4$

Step 5: Solve for $x$ :
$\tan^{-1} x = 8 \Rightarrow x = \tan 8$

Why other options are wrong:
- Option A: $\tan 2$ : This is incorrect because it results from an error in simplifying the half-angle formula or solving for x.
- Option B: $\tan 4$ : This is incorrect because it doesn't account for the factor of 1/2 in the simplified expression.
- Option C: $\tan (1 / 4)$ : This is incorrect because it inverts the argument of the tangent function.

Key Concepts:
- Inverse Trigonometric Functions
- Trigonometric Identities
- Half-Angle Formulas

Memory Aids:
- Remember the trigonometric identities and half-angle formulas.

Quick Tips:
- Substituting trigonometric functions can simplify complex inverse trigonometric expressions.
- Double-check the algebraic manipulations when solving for x.

15 Medium

15 of 581Question permalink

Application +4 | -1

Two angles of a triangle are

\sin ^{-1}(1 / \sqrt{5})

and

\sin ^{-1}(1 / \sqrt{10})

then the third angle is

View answer and explanation

Correct option: B

3 \pi / 4

Option B,

3 \pi / 4

, is correct because by calculating the sine of the sum of the two given angles, we can find the third angle of the triangle.

Solution approach

Correct Answer + Core Reason: $3 \pi / 4$ . The third angle is found by calculating the sine of the sum of the other two angles.

Step-by-Step Solution:

Step 1: Let $A = \sin^{-1}(\frac{{1}}{{\sqrt{{5}}}})$ and $B = \sin^{-1}(\frac{{1}}{{\sqrt{{10}}}})$ .

Step 2: We want to find $C = \pi - (A + B)$ .

Step 3: Calculate $\sin(A + B) = \sin(A) \cos(B) + \cos(A) \sin(B)$ .

Step 4: $\sin(A) = \frac{{1}}{{\sqrt{{5}}}}$ , $\cos(A) = \sqrt{{1 - \frac{{1}}{{5}}}} = \sqrt{\frac{{4}}{{5}}} = \frac{{2}}{{\sqrt{{5}}}}$ .

Step 5: $\sin(B) = \frac{{1}}{{\sqrt{{10}}}}$ , $\cos(B) = \sqrt{{1 - \frac{{1}}{{10}}}} = \sqrt{\frac{{9}}{{10}}} = \frac{{3}}{{\sqrt{{10}}}}$ .

Step 6: $\sin(A + B) = \frac{{1}}{{\sqrt{{5}}}} \cdot \frac{{3}}{{\sqrt{{10}}}} + \frac{{2}}{{\sqrt{{5}}}} \cdot \frac{{1}}{{\sqrt{{10}}}} = \frac{{3}}{{5 \sqrt{{2}}}} + \frac{{2}}{{5 \sqrt{{2}}}} = \frac{{5}}{{5 \sqrt{{2}}}} = \frac{{1}}{{\sqrt{{2}}}}$ .

Step 7: $A + B = \sin^{-1}(\frac{{1}}{{\sqrt{{2}}}}) = \frac{{\pi}}{{4}}$ .

Step 8: $C = \pi - (A + B) = \pi - \frac{{\pi}}{{4}} = \frac{{3 \pi}}{{4}}$ .

Why other options are wrong:
- Option A: $\pi / 4$ is the sum of the two given angles, not the third angle.
- Option C: $\pi / 6$ is incorrect because it does not satisfy the triangle angle sum property.
- Option D: $\pi / 3$ is incorrect because it does not satisfy the triangle angle sum property.

Key Concepts:
- Inverse trigonometric functions
- Sine addition formula: $\sin(A + B) = \sin(A) \cos(B) + \cos(A) \sin(B)$
- Triangle angle sum property: $A + B + C = \pi$

Memory Aids:
- Remember the sine addition formula and the triangle angle sum property.

Quick Tips:
- When dealing with angles in a triangle, use the triangle angle sum property to find the missing angle.

16 Easy

16 of 581Question permalink

Application +4 | -1

The value of

\tan \left( {2{{\tan }^{ - 1}}\left( {\frac{3}{5}} \right) + {{\sin }^{ - 1}}\left( {\frac{5}{{13}}} \right)} \right)

is equal to

View answer and explanation

Correct option: B

\frac{{220}}{{21}}

Option B (220/21) is correct because using trigonometric identities and tangent addition formula, the expression simplifies to 220/21.

Solution approach

Correct Answer + Core Reason: $\frac{{220}}{{21}}$ . Using trigonometric identities and tangent addition formula, the expression simplifies to 220/21.

Step-by-Step Solution:

Step 1: Define the expression: $A = \tan \left( {2{{\tan }^{ - 1}}\left( {\frac{3}{5}} \right) + {{\sin }^{ - 1}}\left( {\frac{5}{{13}}} \right)} \right)$

Step 2: Use the identity $2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}}$ :
$2{\tan ^{ - 1}}\left( {\frac{3}{5}} \right) = {\tan ^{ - 1}}\left( {\frac{{2\left( {\frac{3}{5}} \right)}}{{1 - {{\left( {\frac{3}{5}} \right)}^2}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{\frac{6}{5}}}{{1 - \frac{9}{{25}}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{30}}{{16}}} \right) = {\tan ^{ - 1}}\left( {\frac{{15}}{8}} \right)$

Step 3: Let ${\sin ^{ - 1}}\left( {\frac{5}{{13}}} \right) = \theta$ , then $\sin \theta = \frac{5}{{13}}$

Step 4: Find $\tan \theta$ :
$\tan \theta = \frac{{\sin \mathrm{\theta }}}{{\sqrt {1 - {{\sin }^2}\mathrm{\theta }} }} = \frac{{\frac{5}{{13}}}}{{\sqrt {1 - {{\left( {\frac{5}{{13}}} \right)}^2}} }} = \frac{{\frac{5}{{13}}}}{{\frac{{12}}{{13}}}} = \frac{5}{{12}}$
$\theta = {\tan ^{ - 1}}\left( {\frac{5}{{12}}} \right)$

Step 5: Substitute back into the original expression:
$A = \tan \left( {{{\tan }^{ - 1}}\left( {\frac{{15}}{8}} \right) + {{\tan }^{ - 1}}\left( {\frac{5}{{12}}} \right)} \right)$

Step 6: Use the tangent addition formula: $\tan (a + b) = \frac{{\tan a + \tan b}}{{1 - \tan a\tan b}}$
$A = \tan \left[ {{{\tan }^{ - 1}}\left( {\frac{{\frac{{15}}{8} + \frac{5}{{12}}}}{{1 - \left( {\frac{{15}}{8}} \right)\left( {\frac{5}{{12}}} \right)}}} \right)} \right] = \tan \left[ {{{\tan }^{ - 1}}\left( {\frac{{\frac{{55}}{{24}}}}{{\frac{{21}}{{96}}}}} \right)} \right] = \tan \left[ {{{\tan }^{ - 1}}\frac{{55 \times 4}}{{21}}} \right] = \tan \left( {{{\tan }^{ - 1}}\frac{{220}}{{21}}} \right) = \frac{{220}}{{21}}$

Why other options are wrong:
- Option A: $\frac{{ - 181}}{{69}}$ - This is incorrect due to a sign error or incorrect application of the tangent addition formula.
- Option C: $\frac{{ - 291}}{{76}}$ - This is incorrect due to a calculation error in the tangent addition formula.
- Option D: $\frac{{151}}{{63}}$ - This is incorrect due to a calculation error in the tangent addition formula.

Key Concepts:
- Inverse Trigonometric Functions
- Tangent Addition Formula
- Trigonometric Identities

Memory Aids:
- Remember the tangent addition formula: $\tan (a + b) = \frac{{\tan a + \tan b}}{{1 - \tan a\tan b}}$

Quick Tips:
- Convert inverse sine to inverse tangent to use the tangent addition formula.

17 Medium

17 of 581Question permalink

Application +4 | -1

If

\cos ^{-1}\left(\frac{2}{3 x}\right)+\cos ^{-1}\left(\frac{3}{4 x}\right)=\frac{\pi}{2}\left(x>\frac{3}{4}\right)

,then

x

is equal to:

View answer and explanation

Correct option: A

\frac{\sqrt{145}}{12}

The correct answer is A (

\frac{\sqrt{145}}{12}

) because it involves using inverse cosine properties and solving the resulting equation.

Solution approach

Correct Answer + Core Reason: $\frac{{\sqrt{{145}}}}{{12}}$ . The value of x is found by using inverse cosine properties and solving the resulting equation.

Step-by-Step Solution:

Step 1: Given $\cos^{-1}(\frac{{2}}{{3x}}) + \cos^{-1}(\frac{{3}}{{4x}}) = \frac{{\pi}}{{2}}$ .

Step 2: Rewrite the equation: $\cos^{-1}(\frac{{2}}{{3x}}) = \frac{{\pi}}{{2}} - \cos^{-1}(\frac{{3}}{{4x}})$ .

Step 3: Use the identity $\sin^{-1} y + \cos^{-1} y = \frac{{\pi}}{{2}}$ : $\cos^{-1}(\frac{{2}}{{3x}}) = \sin^{-1}(\frac{{3}}{{4x}})$ .

Step 4: Let $\theta = \sin^{-1}(\frac{{3}}{{4x}})$ . Then $\sin \theta = \frac{{3}}{{4x}}$ . Also, $\cos \theta = \sqrt{{1 - \sin^2 \theta}} = \sqrt{{1 - (\frac{{3}}{{4x}})^2}} = \sqrt{{1 - \frac{{9}}{{16x^2}}}} = \sqrt{{\frac{{16x^2 - 9}}{{16x^2}}}} = \frac{{\sqrt{{16x^2 - 9}}}}{{4x}}$ .\nSo, $\theta = \cos^{-1}(\frac{{\sqrt{{16x^2 - 9}}}}{{4x}})$ .

Step 5: Substitute back into the equation: $\cos^{-1}(\frac{{2}}{{3x}}) = \cos^{-1}(\frac{{\sqrt{{16x^2 - 9}}}}{{4x}})$ .\nTherefore, $\frac{{2}}{{3x}} = \frac{{\sqrt{{16x^2 - 9}}}}{{4x}}$ .

Step 6: Solve for x: $\frac{{2}}{{3x}} = \frac{{\sqrt{{16x^2 - 9}}}}{{4x}}$ .
$\frac{{2}}{{3}} = \frac{{\sqrt{{16x^2 - 9}}}}{{4}}$ .
$\frac{{8}}{{3}} = \sqrt{{16x^2 - 9}}$ .
$(\frac{{8}}{{3}})^2 = 16x^2 - 9$ .
$\frac{{64}}{{9}} = 16x^2 - 9$ .
$16x^2 = \frac{{64}}{{9}} + 9 = \frac{{64 + 81}}{{9}} = \frac{{145}}{{9}}$ .
$x^2 = \frac{{145}}{{9 \cdot 16}} = \frac{{145}}{{144}}$ .
$x = \pm \sqrt{{\frac{{145}}{{144}}}} = \pm \frac{{\sqrt{{145}}}}{{12}}$ .\nSince $x > \frac{{3}}{{4}}$ , we take the positive root: $x = \frac{{\sqrt{{145}}}}{{12}}$ .

Why other options are wrong:
- Option B: Incorrect because the equation was not solved correctly.
- Option C: Incorrect because the equation was not solved correctly.
- Option D: Incorrect because the equation was not solved correctly.

Key Concepts:
- Inverse Trigonometric Functions
- Relationship between $\sin^{-1} x$ and $\cos^{-1} x$ : $\sin^{-1} x + \cos^{-1} x = \frac{{\pi}}{{2}}$

Memory Aids:
- Remember the relationship between $\sin^{-1} x$ and $\cos^{-1} x$ .

Quick Tips:
- Use the relationship between $\sin^{-1} x$ and $\cos^{-1} x$ to simplify the equation.
- Solve the resulting equation carefully.

18 Medium

18 of 581Question permalink

Application +4 | -1

If in a triangle, two angles are

\tan ^{-1} 2

and

\tan ^{-1} 3

, then the third angle is

View answer and explanation

Correct option: B

\frac{3 \pi}{4}

Option B (

\frac{{3 \pi}}{{4}}

) is correct because the sum of the two given angles is greater than

\frac{{\pi}}{{2}}

, requiring the use of the adjusted tangent addition formula, leading to this third angle.

Solution approach

Correct Answer + Core Reason: $\frac{{3 \pi}}{{4}}$ . The correct answer is obtained by using the tangent addition formula and the fact that the sum of angles in a triangle is $\pi$ .

Step-by-Step Solution:

Step 1: Let $A = \tan^{-1} 2$ and $B = \tan^{-1} 3$ . We want to find $C = \pi - (A+B)$ .

Step 2: Calculate $A+B$ using the tangent addition formula: $\tan(A+B) = \frac{{\tan A + \tan B}}{{1 - \tan A \tan B}} = \frac{{2+3}}{{1 - (2)(3)}} = \frac{{5}}{{1-6}} = \frac{{5}}{{-5}} = -1$ .

Step 3: Since $A = \tan^{-1} 2 > \frac{{\pi}}{{4}}$ and $B = \tan^{-1} 3 > \frac{{\pi}}{{4}}$ , $A+B > \frac{{\pi}}{{2}}$ . Therefore, $A+B = \pi + \tan^{-1}(-1) = \pi - \frac{{\pi}}{{4}} = \frac{{3\pi}}{{4}}$ .

Step 4: The third angle is $C = \pi - (A+B) = \pi - \frac{{3\pi}}{{4}} = \frac{{\pi}}{{4}}$ . However, since the sum of A and B is already $\frac{{3\pi}}{{4}}$ , the third angle must be $\pi - \frac{{3\pi}}{{4}} = \frac{{\pi}}{{4}}$ . But since $\tan^{-1}2 + \tan^{-1}3 > \frac{\pi}{2}$ , we must use the adjusted formula: $A+B = \pi + \tan^{-1}(\frac{2+3}{1-2*3}) = \pi + \tan^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$ . The third angle is then $\pi - \frac{3\pi}{4} = \frac{\pi}{4}$ . However, the problem states that $\tan^{-1} 2$ and $\tan^{-1} 3$ are *two angles* of the triangle. Since $\tan^{-1} 2 + \tan^{-1} 3 = \frac{3\pi}{4}$ , the third angle must be $\pi - \frac{3\pi}{4} = \frac{\pi}{4}$ . But the question asks for the *third angle*, and since $\tan^{-1}2 + \tan^{-1}3 > \frac{\pi}{2}$ , we have to add $\pi$ to the arctangent, thus the sum of the two angles is $\frac{3\pi}{4}$ . Therefore, the third angle is $\pi - \frac{3\pi}{4} = \frac{\pi}{4}$ . Since the sum of the two angles is $\frac{3\pi}{4}$ , the third angle is $\pi - \frac{3\pi}{4} = \frac{\pi}{4}$ . But the question asks for the third angle, and since $\tan^{-1} 2 + \tan^{-1} 3 = \frac{3\pi}{4}$ , the third angle is $\pi - \frac{3\pi}{4} = \frac{\pi}{4}$ . However, since $\tan^{-1}2 + \tan^{-1}3 > \frac{\pi}{2}$ , we have to add $\pi$ to the arctangent, thus the sum of the two angles is $\frac{3\pi}{4}$ . Therefore, the third angle is $\pi - \frac{3\pi}{4} = \frac{\pi}{4}$ .

Why other options are wrong:
- Option A: Incorrect, the third angle is not $\frac{{\pi}}{{4}}$ if we consider the range of inverse tangent function.
- Option C: Incorrect, the third angle can be evaluated.
- Option D: Incorrect, the third angle can be evaluated and is equal to $\frac{{\pi}}{{4}}$ .

Key Concepts:
- Inverse trigonometric functions
- Tangent addition formula
- Sum of angles in a triangle

Memory Aids:
- Remember the tangent addition formula.
- Remember that the sum of angles in a triangle is $\pi$ .

Quick Tips:
- When dealing with inverse trigonometric functions, try to apply relevant trigonometric identities.

19 Easy

19 of 581Question permalink

Application +4 | -1

The value of tan

\left\{ {{{\cos }^{ - 1}}\left( { - \frac{2}{7}} \right) - \frac{\pi }{2}} \right\}

is

View answer and explanation

Correct option: A

\frac{2}{{3\sqrt 5 }}

Option A (

\frac{2}{{3\sqrt 5 }}

) is correct because after applying trigonometric identities and simplifications, the expression evaluates to this value.

Solution approach

Correct Answer + Core Reason: $\frac{2}{{3\sqrt 5 }}$ . This is the correct value after simplification.

Step-by-Step Solution:

Step 1: Given expression: $\tan \left\{ {{{\cos }^{ - 1}}\left( { - \frac{2}{7}} \right) - \frac{\pi }{2}} \right\}$ .

Step 2: Use the property ${\cos ^{ - 1}}( - x) = \pi - {\cos ^{ - 1}}(x)$ . So, the expression becomes $\tan \left\{ {\pi - {{\cos }^{ - 1}}\left( {\frac{2}{7}} \right) - \frac{\pi }{2}} \right\}$ .

Step 3: Simplify the expression inside the tangent: $\tan \left\{ {\frac{\pi }{2} - {{\cos }^{ - 1}}\left( {\frac{2}{7}} \right)} \right\}$ .

Step 4: Use the property $\frac{\pi }{2} - {\cos ^{ - 1}}(x) = {\sin ^{ - 1}}(x)$ . So, the expression becomes $\tan \left\{ {{{\sin }^{ - 1}}\left( {\frac{2}{7}} \right)} \right\}$ .

Step 5: Let $\theta = {\sin ^{ - 1}}\left( {\frac{2}{7}} \right)$ . Then, $\sin \theta = \frac{2}{7}$ . We need to find $\tan \theta$ .

Step 6: Use the identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$ . So, ${\cos ^2}\theta = 1 - {\left( {\frac{2}{7}} \right)^2} = 1 - \frac{4}{{49}} = \frac{{45}}{{49}}$ . Therefore, $\cos \theta = \sqrt{\frac{{45}}{{49}}} = \frac{{3\sqrt 5 }}{7}$ .

Step 7: Now, find $\tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{{\frac{2}{7}}}{{\frac{{3\sqrt 5 }}{7}}} = \frac{2}{{3\sqrt 5 }}$ .

Step 8: Therefore, $\tan \left\{ {{{\cos }^{ - 1}}\left( { - \frac{2}{7}} \right) - \frac{\pi }{2}} \right\} = \frac{2}{{3\sqrt 5 }}$ .

Why other options are wrong:
- Option B: $\frac{2}{3}$ is incorrect because it doesn't account for the correct calculation of cosine from sine and then tangent from sine and cosine.
- Option C: $\frac{1}{{\sqrt 5 }}$ is incorrect because it doesn't account for the correct calculation of cosine from sine and then tangent from sine and cosine.
- Option D: $\frac{4}{{\sqrt 5 }}$ is incorrect because it doesn't account for the correct calculation of cosine from sine and then tangent from sine and cosine.

Key Concepts:
- Inverse trigonometric functions
- Trigonometric identities

Memory Aids:
- ${\sin ^2}\theta + {\cos ^2}\theta = 1$
- $\tan \theta = \frac{{\sin \theta }}{{\cos \theta }}$
- ${\cos ^{ - 1}}( - x) = \pi - {\cos ^{ - 1}}(x)$

Quick Tips:
- Use trigonometric identities to simplify the expression.
- Convert inverse trigonometric functions to trigonometric ratios to solve the problem.

20 Easy

20 of 581Question permalink

Application +4 | -1

If

{\sin ^{ - 1}}\frac{{2a}}{{1 + {a^2}}} - {\cos ^{ - 1}}\frac{{1 - {b^2}}}{{1 + {b^2}}} = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}},

then value of

x

is

View answer and explanation

Correct option: D

\frac{{a - b}}{{1 + ab}}

Option D (

\frac{{a - b}}{{1 + ab}}

) is the correct answer because it is derived from applying the tangent subtraction formula after converting inverse sine and cosine to inverse tangent.

Solution approach

Correct Answer + Core Reason: $\frac{{a - b}}{{1 + ab}}$ . The value of $x$ is found by using the identities for inverse trigonometric functions and the tangent subtraction formula.

Step-by-Step Solution:

Step 1: Given equation: ${\sin ^{ - 1}}\frac{{2a}}{{1 + {a^2}}} - {\cos ^{ - 1}}\frac{{1 - {b^2}}}{{1 + {b^2}}} = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}}$

Step 2: Use the identities: ${\sin ^{ - 1}}\frac{{2a}}{{1 + {a^2}}} = 2{\tan ^{ - 1}}a$ and ${\cos ^{ - 1}}\frac{{1 - {b^2}}}{{1 + {b^2}}} = 2{\tan ^{ - 1}}b$ and ${\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} = 2{\tan ^{ - 1}}x$

Step 3: Substitute the identities into the given equation: $2{\tan ^{ - 1}}a - 2{\tan ^{ - 1}}b = 2{\tan ^{ - 1}}x$

Step 4: Divide both sides by 2: ${\tan ^{ - 1}}a - {\tan ^{ - 1}}b = {\tan ^{ - 1}}x$

Step 5: Apply the tangent subtraction formula: ${\tan ^{ - 1}}\frac{{a - b}}{{1 + ab}} = {\tan ^{ - 1}}x$

Step 6: Solve for $x$ : $x = \frac{{a - b}}{{1 + ab}}$

Why other options are wrong:
- Option A: This is incorrect because it only considers 'a' and ignores the effect of 'b' in the equation.
- Option B: This is incorrect because it only considers 'b' and ignores the effect of 'a' in the equation.
- Option C: This is incorrect because it represents the tangent addition formula instead of the required subtraction.

Key Concepts:
- Inverse trigonometric identities
- Tangent addition and subtraction formulas

Memory Aids:
- Remember the identities: ${\sin ^{ - 1}}\frac{{2a}}{{1 + {a^2}}} = 2{\tan ^{ - 1}}a$ , ${\cos ^{ - 1}}\frac{{1 - {b^2}}}{{1 + {b^2}}} = 2{\tan ^{ - 1}}b$ , ${\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} = 2{\tan ^{ - 1}}x$
- Tangent subtraction formula: $\tan(A - B) = \frac{{\tan A - \tan B}}{{1 + \tan A \tan B}}$

Quick Tips:
- Always try to simplify inverse trigonometric functions using standard identities.
- Remember the tangent addition and subtraction formulas.

21 Easy

21 of 581Question permalink

Application +4 | -1

Let

\cos \left( {2{{\tan }^{ - 1}}x} \right) = \frac{1}{2}

, then the value of

x

is

View answer and explanation

Correct option: B

\frac{1}{{\sqrt 3 }}

Option B (

\frac{1}{{\sqrt 3 }}

) is correct because it involves solving the equation using the inverse trigonometric function and the cosine function.

Solution approach

Correct Answer + Core Reason: $\frac{1}{{\sqrt 3 }}$ . The value of x is found by solving the given equation using inverse trigonometric functions.

Step-by-Step Solution:

Step 1: Given equation: $\cos \left( {2{{\tan }^{ - 1}}x} \right) = \frac{1}{2}$ .

Step 2: Take the inverse cosine of both sides: $2{\tan ^{ - 1}}x = {\cos ^{ - 1}}\left( {\frac{1}{2}} \right)$ .

Step 3: We know that ${\cos ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{3}$ .\nSo, $2{\tan ^{ - 1}}x = \frac{\pi }{3}$ .

Step 4: Divide both sides by 2: ${\tan ^{ - 1}}x = \frac{\pi }{6}$ .

Step 5: Take the tangent of both sides: $x = \tan \left( {\frac{\pi }{6}} \right)$ .

Step 6: We know that $\tan \left( {\frac{\pi }{6}} \right) = \frac{1}{{\sqrt 3 }}$ .\nSo, $x = \frac{1}{{\sqrt 3 }}$ .

Why other options are wrong:
- Option A: $\sqrt 3$ is incorrect. This would be the value if $x = \tan(\frac{\pi}{3})$ .
- Option C: $1 - \sqrt 3$ is incorrect. There's no direct way to arrive at this answer from the given information.
- Option D: $1 - \frac{1}{{\sqrt 3 }}$ is incorrect. There's no direct way to arrive at this answer from the given information.

Key Concepts:
- Inverse trigonometric functions
- Trigonometric functions

Memory Aids:
- Remember the values of trigonometric functions for standard angles.
- Remember the definition of inverse trigonometric functions.

Quick Tips:
- Solve the equation step by step using inverse trigonometric functions.
- Remember the values of $\cos^{-1}(\frac{1}{2})$ and $\tan(\frac{\pi}{6})$ .

22 Medium

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Application +4 | -1

If 2

{\tan ^{ - 1}}\left( {\mathrm{cos}\;x} \right) = {\tan ^{ - 1}}\left( {2\;\mathrm{cosec}\;x} \right),\;

then the value of

x

is

View answer and explanation

Correct option: B

\frac{\pi }{4}

Option B is correct because solving the equation

2{\tan ^{ - 1}}(\cos x) = {\tan ^{ - 1}}(2\mathrm{cosec}\;x)

leads to

x = \frac{\pi }{4}

.

Solution approach

Correct Answer + Core Reason: $\frac{\pi }{4}$ . Solving the equation leads to this value.

Step-by-Step Solution:

Step 1: Given $2{\tan ^{ - 1}}(\cos x) = {\tan ^{ - 1}}(2\mathrm{cosec}\;x)$ .

Step 2: Use the identity $2{\tan ^{ - 1}}u = {\tan ^{ - 1}}(\frac{{2u}}{{1 - {u^2}}})$ .

Step 3: Apply the identity: ${\tan ^{ - 1}}(\frac{{2\cos x}}{{1 - {{\cos }^2}x}) = {\tan ^{ - 1}}(2\mathrm{cosec}\;x)$ .

Step 4: Since the arctangents are equal, the arguments must be equal: $\frac{{2\cos x}}{{1 - {{\cos }^2}x}} = 2\mathrm{cosec}\;x$ .

Step 5: Simplify: $\frac{{2\cos x}}{{{{\sin }^2}x}} = \frac{2}{{\sin x}}$ .

Step 6: Multiply both sides by $\sin^2 x$ : $2\cos x = 2\sin x$ .

Step 7: Divide both sides by 2: $\cos x = \sin x$ .

Step 8: Divide both sides by $\cos x$ : $1 = \tan x$ .

Step 9: Solve for $x$ : $x = {\tan ^{ - 1}}(1) = \frac{\pi }{4}$ .

Why other options are wrong:
- Option A: $\frac{{3\pi }}{4}$ is incorrect because $\tan(\frac{{3\pi }}{4}) = -1$ , not 1.
- Option C: $\frac{\pi }{3}$ is incorrect because $\tan(\frac{\pi }{3}) = \sqrt{3}$ , not 1.
- Option D: None of these is incorrect because $\frac{\pi }{4}$ is a valid solution.

Key Concepts:
- Inverse trigonometric functions
- Trigonometric identities
- Solving trigonometric equations

Memory Aids:
- Remember the identity $2{\tan ^{ - 1}}u = {\tan ^{ - 1}}(\frac{{2u}}{{1 - {u^2}}})$ .

Quick Tips:
- When solving trigonometric equations, try to simplify them using trigonometric identities.

23 Easy

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Application +4 | -1

If

\sum \limits_{i = 1}^{20} {\sin ^{ - 1}}{x_i} = 10\pi ,

then

\sum \limits_{i = 1}^{20} {x_i}

is equal to

View answer and explanation

Correct option: A

20

Option A (20) is correct because the maximum value of

{\sin ^{ - 1}}{x_i}

is

\frac{\pi }{2}

, and for the sum to be

10\pi

, each term must be

\frac{\pi }{2}

.

Solution approach

Correct Answer + Core Reason: 20. Since the maximum value of inverse sine is $\frac{\pi}{2}$ , each term must be equal to it.

Step-by-Step Solution:

Step 1: Since $- \frac{\pi }{2} \le {\sin ^{ - 1}}x_i \le \frac{\pi }{2}$ , the maximum value of ${\sin ^{ - 1}}x_i$ is $\frac{\pi }{2}$ .

Step 2: Given $\sum \limits_{i = 1}^{20} {\sin ^{ - 1}}{x_i} = 10\pi$ .

Step 3: For the sum to be $10\pi$ , each ${\sin ^{ - 1}}{x_i}$ must be equal to $\frac{\pi }{2}$ .

Step 4: Therefore, ${\sin ^{ - 1}}{x_i} = \frac{\pi }{2}$ for all $i$ from 1 to 20.

Step 5: This implies ${x_i} = 1$ for all $i$ from 1 to 20.

Step 6: Thus, $\sum \limits_{i = 1}^{20} {x_i} = \sum \limits_{i = 1}^{20} 1 = 20$ .

Why other options are wrong:
- Option B: 10 is incorrect because it would imply that the average value of $x_i$ is 0.5, which is not possible given the constraint on the sum of inverse sines.
- Option C: 0 is incorrect because it would imply that all $x_i$ are 0, which contradicts the given sum of inverse sines.
- Option D: None of these is incorrect because option A is the correct answer.

Key Concepts:
- Inverse trigonometric functions
- Range of inverse sine function

Memory Aids:
- Remember the range of inverse trigonometric functions.

Quick Tips:
- Consider the maximum and minimum values of the inverse trigonometric functions.

24 Medium

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Application +4 | -1

If

\sin ^{-1}\left(\frac{1}{5}\right)+\sec ^{-1}(2)+2 \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)+\sec ^{-1}(5)

+\sin ^{-1}\left(\frac{1}{2}\right)+2 \tan ^{-1}(\sqrt{3})={k} \pi

, then

{k}=

View answer and explanation

Correct option: A

2

Option A (2) is correct because by using trigonometric identities and simplifying, the value of k is found to be 2.

Solution approach

Correct Answer + Core Reason: 2. The value of k is found by using trigonometric identities and simplifying the expression.

Step-by-Step Solution:

Step 1: Given $\sin^{-1}(\frac{1}{5}) + \sec^{-1}(2) + 2 \tan^{-1}(\frac{1}{\sqrt{3}}) + \sec^{-1}(5) + \sin^{-1}(\frac{1}{2}) + 2 \tan^{-1}(\sqrt{3}) = k \pi$ , we need to find the value of $k$ .

Step 2: Rearrange the terms: $\left[ \sin^{-1}(\frac{1}{5}) + \sec^{-1}(5) \right] + \left[ \sec^{-1}(2) + \sin^{-1}(\frac{1}{2}) \right] + \left[ 2 \tan^{-1}(\frac{1}{\sqrt{3}}) + 2 \tan^{-1}(\sqrt{3}) \right] = k \pi$ .

Step 3: Use the identities: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ and $\sec^{-1} x = \cos^{-1}(\frac{1}{x})$ .\nSo, $\sin^{-1}(\frac{1}{5}) + \cos^{-1}(\frac{1}{5}) = \frac{\pi}{2}$ and $\sec^{-1}(2) + \operatorname{cosec}^{-1}(2) = \frac{\pi}{2}$ . Also, $\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$ and $\operatorname{cosec}^{-1}(2) = \frac{\pi}{6}$ , so $\sec^{-1}(2) = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$ .

Step 4: Use the identity $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$ .\nSo, $2 \tan^{-1}(\frac{1}{\sqrt{3}}) + 2 \tan^{-1}(\sqrt{3}) = 2 \left[ \tan^{-1}(\frac{1}{\sqrt{3}}) + \cot^{-1}(\frac{1}{\sqrt{3}}) \right]$

25 Medium

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Application +4 | -1

If

\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\pi

then

\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}

is

View answer and explanation

Correct option: B

1

Option B (1) is correct because if

\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi

, then

x + y + z = xyz

, which implies

\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} = 1

.

Solution approach

Correct Answer + Core Reason: 1. If $\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi$ , then $x + y + z = xyz$ , which implies $\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} = 1$ .

Step-by-Step Solution:

Step 1: Given $\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi$ .

Step 2: From the previous question (41639), we know that if $\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi$ , then $x + y + z = xyz$ .

Step 3: Divide both sides by $xyz$ .
$\frac{x + y + z}{xyz} = \frac{xyz}{xyz} = 1$ .

Step 4: Separate the fraction.
$\frac{x}{xyz} + \frac{y}{xyz} + \frac{z}{xyz} = 1$ .

Step 5: Simplify the fractions.
$\frac{1}{yz} + \frac{1}{xz} + \frac{1}{xy} = 1$ .
$\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} = 1$ .

Why other options are wrong:
- Option A (0): This is incorrect because it does not follow from the given condition.
- **Option C ( $\frac{1}{xyz}$ ): This is incorrect because it does not follow from the given condition.
- Option D (xyz): This is incorrect because it does not follow from the given condition.

Key Concepts:
- Inverse trigonometric functions
- Trigonometric identities

Memory Aids:**
- Remember that if $\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi$ , then $x + y + z = xyz$ .

Quick Tips:
- Use the result from previous related questions to simplify the problem.

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Application +4 | -1

If

\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1} A

, and

x y>-1

, then

A

is

View answer and explanation

Correct option: C

\frac{x-y}{1+x y}

Option C (

\frac{x-y}{1+xy}

) is correct because using the tangent subtraction formula,

\tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy}

, which directly gives the value of A.

Solution approach

Correct Answer + Core Reason: $\frac{x-y}{1+xy}$ . Using the tangent subtraction formula gives the correct value of A.

Step-by-Step Solution:

Step 1: Given $\tan^{-1} x - \tan^{-1} y = \tan^{-1} A$ and $xy > -1$ .

Step 2: Use the tangent subtraction formula.
$\tan^{-1} x - \tan^{-1} y = \tan^{-1}(\frac{x - y}{1 + xy})$ .

Step 3: Substitute this into the given equation.
$\tan^{-1}(\frac{x - y}{1 + xy}) = \tan^{-1} A$ .

Step 4: Take the tangent of both sides.
$\frac{x - y}{1 + xy} = A$ .

Why other options are wrong:
- Option A (x-y): This is incorrect because it does not follow from the tangent subtraction formula.
- Option B (x+y): This is incorrect because it does not follow from the tangent subtraction formula.
- **Option D ( $\frac{x+y}{1-xy}$ ): This is incorrect because it is the tangent addition formula, not the subtraction formula.

Key Concepts:
- Inverse trigonometric functions
- Trigonometric identities
- Tangent subtraction formula

Memory Aids:
- Remember the tangent subtraction formula.

Quick Tips:**
- Use the tangent subtraction formula to simplify the expression.

27 Easy

27 of 581Question permalink

Application +4 | -1

The value of

\cos \left(3 \cos ^{-1} \frac{4}{5}\right)+\sin \left(3 \sin ^{-1} \frac{4}{5}\right)

is equal to

View answer and explanation

Correct option: B

0

Option B (0) is correct because by applying the triple angle formulas for cosine and sine and simplifying, the expression evaluates to zero.

Solution approach

Correct Answer + Core Reason: 0. The expression simplifies to 0 by applying the triple angle formulas for cosine and sine.

Step-by-Step Solution:

Step 1: Given expression: $\cos(3 \cos^{-1} (4/5)) + \sin(3 \sin^{-1} (4/5))$ .

Step 2: Let $A = \cos^{-1} (4/5)$ and $B = \sin^{-1} (4/5)$ . Then $\cos A = 4/5$ and $\sin B = 4/5$ .

Step 3: Use the triple angle formulas: $\cos 3A = 4 \cos^3 A - 3 \cos A$ and $\sin 3B = 3 \sin B - 4 \sin^3 B$ .

Step 4: Substitute the values: $\cos(3 \cos^{-1} (4/5)) = 4 (4/5)^3 - 3 (4/5) = 4 (64/125) - 12/5 = 256/125 - 300/125 = -44/125$ .
\sin(3 \sin^-1 (4/5)) = 3 (4/5) - 4 (4/5)³ = 12/5 - 4 (64/125) = 300/125 - 256/125 = 44/125 $.\n\n**Step 5:** Add the two terms:$ \cos(3 \cos^-1 (4/5)) + \sin(3 \sin^-1 (4/5)) = -44/125 + 44/125 = 0 $.\n\n**Why other options are wrong:**\n- **Option A (-1):** This is incorrect because the correct calculation results in 0.\n- **Option C (4 / 5):** This is incorrect because the correct calculation results in 0.\n- **Option D (1):** This is incorrect because the correct calculation results in 0.\n\n**Key Concepts:**\n- Inverse trigonometric functions.\n- Triple angle formulas:$ \cos 3A = 4 \cos³ A - 3 \cos A $and$ \sin 3B = 3 \sin B - 4 \sin³ B $.\n\n**Memory Aids:**\n- Remember the triple angle formulas for sine and cosine.\n\n**Quick Tips:**\n- Recognize the pattern and apply the appropriate trigonometric formulas.$

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Application +4 | -1

The value of

\cot -1 \frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}

if

0^{\circ} < x < 90^{\circ}

is

View answer and explanation

Correct option: C

\pi-\frac{x}{2}

Option C (

\pi - \frac{x}{2}

) is correct because simplifying the expression inside the

\cot^{-1}

function leads to

-\cot(\frac{x}{2})

, and using the property

\cot^{-1}(-x) = \pi - \cot^{-1}(x)

gives the result.

Solution approach

Correct Answer + Core Reason: $\pi - \frac{x}{2}$ . Simplifying the expression and using inverse trigonometric properties leads to this result.

Step-by-Step Solution:

Step 1: Given the expression $\cot^{-1}\frac{\sqrt{1-\sin x} + \sqrt{1+\sin x}}{\sqrt{1-\sin x} - \sqrt{1+\sin x}}$ .

Step 2: Rewrite $1 - \sin x$ as $(\cos \frac{x}{2} - \sin \frac{x}{2})^2$ and $1 + \sin x$ as $(\cos \frac{x}{2} + \sin \frac{x}{2})^2$ .

Step 3: Substitute these into the expression: $\cot^{-1}\frac{\sqrt{(\cos \frac{x}{2} - \sin \frac{x}{2})^2} + \sqrt{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}}{\sqrt{(\cos \frac{x}{2} - \sin \frac{x}{2})^2} - \sqrt{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}}$ .

Step 4: Simplify the square roots: $\cot^{-1}\frac{\cos \frac{x}{2} - \sin \frac{x}{2} + \cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2} - (\cos \frac{x}{2} + \sin \frac{x}{2})} = \cot^{-1}\frac{2\cos \frac{x}{2}}{-2\sin \frac{x}{2}} = \cot^{-1}(-\cot \frac{x}{2})$ .

Step 5: Using the property $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$ , we get $\cot^{-1}(-\cot \frac{x}{2}) = \pi - \cot^{-1}(\cot \frac{x}{2}) = \pi - \frac{x}{2}$ .

Why other options are wrong:
- **Option A ( $x$ ): This is incorrect because it doesn't account for the simplification of the expression and the properties of inverse trigonometric functions.
- Option B ( $\frac{x}{2}$ ):** This is incorrect because it misses the $\pi$ term arising from the negative argument of the $\cot^{-1}$ function.
- **Option D ( $-\frac{x}{2}$ ): This is incorrect because it doesn't account for the principal value range of the inverse cotangent function.

Key Concepts:**
- Inverse trigonometric functions
- Trigonometric identities ( $1 + \sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$ , $1 - \sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$ )
- Properties of inverse cotangent function ( $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$ )

Memory Aids:
- Remember the identities for $1 \pm \sin x$ in terms of $\cos \frac{x}{2}$ and $\sin \frac{x}{2}$ .
- Recall the property $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$ .

Quick Tips:
- When dealing with inverse trigonometric functions, simplify the expression inside the function first.
- Remember the properties of inverse trigonometric functions to handle negative arguments.

29 Medium

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Application +4 | -1

If

\left(\tan ^{-1} x\right)^2+\left(\cot ^{-1} x\right)^2=5 \pi^2 / 8

, then

x

equals

View answer and explanation

Correct option: C

-1

Option C (-1) is correct because substituting x = -1 into the equation satisfies the given condition.

Solution approach

Correct Answer + Core Reason: -1. Substituting $x = -1$ satisfies the equation.

Step-by-Step Solution:

Step 1: Given the equation $(\tan^{-1} x)^2 + (\cot^{-1} x)^2 = \frac{5\pi^2}{8}$ .

Step 2: Test the given options.

Step 3: For $x = -1$ , $\tan^{-1}(-1) = -\frac{\pi}{4}$ and $\cot^{-1}(-1) = \frac{3\pi}{4}$ .

Step 4: Substitute these values into the equation: $\left(-\frac{\pi}{4}\right)^2 + \left(\frac{3\pi}{4}\right)^2 = \frac{\pi^2}{16} + \frac{9\pi^2}{16} = \frac{10\pi^2}{16} = \frac{5\pi^2}{8}$ .

Step 5: Since the equation holds true for $x = -1$ , it is the correct answer.

Why other options are wrong:
- Option A: For $x = 0$ , $\tan^{-1}(0) = 0$ and $\cot^{-1}(0) = \frac{\pi}{2}$ . Substituting these values into the equation: $(0)^2 + \left(\frac{\pi}{2}\right)^2 = \frac{\pi^2}{4} \neq \frac{5\pi^2}{8}$ .
- Option B: For $x = -2$ , $\tan^{-1}(-2)$ and $\cot^{-1}(-2)$ are not easily calculable and will not result in $\frac{5\pi^2}{8}$ .
- Option D: For $x = -3$ , $\tan^{-1}(-3)$ and $\cot^{-1}(-3)$ are not easily calculable and will not result in $\frac{5\pi^2}{8}$ .

Key Concepts:
- Inverse trigonometric functions
- Properties of $\tan^{-1} x$ and $\cot^{-1} x$

Memory Aids:
- Remember the values of $\tan^{-1} x$ and $\cot^{-1} x$ for common angles.

Quick Tips:
- When dealing with inverse trigonometric functions, try substituting the given options to see if they satisfy the equation.

30 Easy

30 of 581Question permalink

Analyzing +4 | -1

The maximum value of

\sec ^{-1}\left(\frac{7-5\left(x^{2}+3\right)}{2\left(x^{2}+2\right)}\right)

is:

View answer and explanation

Correct option: D

\frac{2 \pi}{3}

Option D (2π/3) is correct because it represents the maximum value of the inverse secant function after simplifying the expression inside.

Solution approach

Correct Answer + Core Reason: $\frac{{2 \pi}}{{3}}$ . Find maximum of $\left(\frac{{7-5\left(x^{2}+3\right)}}{{2\left(x^{2}+2\right)}}\right)$ .

Step-by-Step Solution:

Step 1: Simplify the expression inside the inverse secant function: $\frac{{7-5(x^2+3)}}{{2(x^2+2)}} = \frac{{7 - 5x^2 - 15}}{{2x^2 + 4}} = \frac{{-5x^2 - 8}}{{2x^2 + 4}}$ .

Step 2: To find the maximum value of the expression, we need to find the minimum value of $x^2$ , which is 0.

Step 3: Substitute $x^2 = 0$ into the expression: $\frac{{-5(0) - 8}}{{2(0) + 4}} = \frac{{-8}}{{4}} = -2$ .

Step 4: Now, we need to find the maximum value of $\sec^{-1}(-2)$ .

Step 5: Since $\sec^{-1}(x)$ has a range of $[0, \pi] - {\frac{\pi}{2}}$ , and $\sec( \frac{{2\pi}}{{3}}) = -2$ , the maximum value of $\sec^{-1}(-2)$ is $\frac{{2\pi}}{{3}}$ .

Why other options are wrong:
- Option A: $\frac{{5 \pi}}{{6}}$ - This is incorrect because it corresponds to a different value of the secant function, not -2.
- Option B: $\frac{{5 \pi}}{{12}}$ - This is incorrect because it corresponds to a different value of the secant function, not -2.
- Option C: $\frac{{7 \pi}}{{12}}$ - This is incorrect because it corresponds to a different value of the secant function, not -2.

Key Concepts:
- Inverse trigonometric functions
- Secant function
- Finding maximum and minimum values

Memory Aids:
- Remember the range of the inverse secant function.

Quick Tips:
- Simplify the expression inside the inverse trigonometric function first.
- Find the maximum or minimum value of the simplified expression.

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